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Math Help - Setting up systems of equations

  1. #1
    m58
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    Setting up systems of equations with several variables

    Hello,

    We simply have to set up a system of equations for homework in class, NOT solve it. However, I'm not sure I set up everything correctly.

    Many small shops closed down their respective cafes in the 1970s. Statistics for 1970, 1975, and 1977 show that a total of 187000 shops closed down their cafes during those years. 1975 was the year with the most cafe closures; only 15000 fewer cafes closed down than in 1970 and 1977 combined. 1970 was the year when the least number of cafes closed; 45000 fewer cafes closed down in 1970 than in 1977. How many shops closed down their cafes in each of the years?

    x= cafes closed in 1970
    y= cafes closed in 1975
    z= cafes closed in 1977

    x + y + z = 187,000
    y = (x + z) - 15,000 -> (x + z) - 15,000 - y = 0
    x = z - 45,000 -> x - z + 45000 = 0

    Problem 2...

    A bike companies uses three machines: F1, F2, and F3 to make three types of bikes: road bikes, racing bikes, and children's bikes. Each road bike must spend 6 hours on machine F1, 1 hour on machine F2, and 3 hours on machine F3. Each racing bike must spend 4 hours on machine F1, 3 hours on machine F2, and 2 hours on machine F3. The children's bikes require 8 hours on machine F1, 1 hour on machine F2, and 4 hours on machine F3. Each machine is available for 50 hours/week. How many bikes of each type should the company make if it plans to use the machines for all the time they are available? Why is it impossible for the company to use all three machines for all the available time? Would it be possible if the company could make non-whole number bikes per week (i.e. 5.3 racing bikes)?

    x= number of road bikes to be manufactured each week
    y= number of racing bikes to be manufactured each week
    z= number of children's bikes to be manufactured each week

    x = 6x + y + 3z
    y= 4x + 3y + 2z
    z= 8x + y + 4z
    x + y + z = 3

    I also said that it would be impossible for the company to make whole number bikes because the matrix from the system of equations would not yield an inverse. It would yield an inverse if non-whole numbers were used, though.
    Last edited by m58; October 8th 2011 at 03:48 PM.
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  2. #2
    m58
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    Re: Setting up systems of equations with several variables

    I was just thinking... do I have to make the equations equal 0, like in y = (x + z) - 15,000 -> (x + z) - 15,000 - y = 0? Or do I just leave them as they are?
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  3. #3
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    Re: Setting up systems of equations with several variables

    Your equations for the first problem are correct from what I can tell, and you can solve it using substitution and rearranging some of the equations (if you like, whether you've been asked to or not; it might look good if you do!). I think it's probably better to leave it in terms of the other variables for what you've been asked though, rather than rearranging to give 0 (which would be better for a polynomial with one variable), i.e:

    x + y + z = 187,000
    y = (x + z) - 15,000
    x = z - 45,000

    Unfortunately, I don't have enough time right now to look at the other problem, but I hope that helps!
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  4. #4
    m58
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    Re: Setting up systems of equations with several variables

    Thanks, FrameofMind! If you have time, could you critique my analysis of the second problem?
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  5. #5
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    Re: Setting up systems of equations with several variables

    Quote Originally Posted by m58 View Post
    x + y + z = 187,000
    y = (x + z) - 15,000 -> (x + z) - 15,000 - y = 0
    You can get y right off the bat:
    x + y + z= 187000 ; arrange 2nd equation this way:
    -x+y - z = -15000
    ============== Add 'em:
    2y = 172000
    Continue...
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  6. #6
    m58
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    Re: Setting up systems of equations with several variables

    Thanks, Wilmer! How does the second problem look, though?
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  7. #7
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    Re: Setting up systems of equations with several variables

    Quote Originally Posted by m58 View Post
    How does the second problem look, though?
    Honestly? TERRIBLE
    Your equations:
    x = 6x + y + 3z : -5x = y + 3z
    y= 4x + 3y + 2z : -2y = 4x + 2z
    z= 8x + y + 4z : -3z = 8x + y
    You can't have a negative number of bikes, right?
    And this: x + y + z = 3
    Why only 3 bikes?

    Anyway, the questions in the problem make no sense to me...very unclear.
    We have this:
    Code:
        F1  F2  F3 
    x   6   1   3
    y   4   3   2
    z   8   1   4
    Maximum bikes in 1 week, with at least one of each kind, taking exactly 50 hours:
    1x + 9y + 1z = 11 bikes; on F1: 6 + 36 + 8 = 50 hours

    Maybe someone else will understand the questions....
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